Comptes Rendus Mathématique

Bruno M. Braga

Javier Alejandro Chávez-Domínguez

On the small scale nonlinear theory of operator spaces

Michiya Mori

Comptes Rendus
Mathématique
Michiya Mori
On the Scottish Book Problem 155 by Mazur and Sternbach
Volume 362 (2024), p. 813-816
Online since: 7 October 2024
https://doi.org/10.5802/crmath.572
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Comptes Rendus. Mathématique
2024, Vol. 362, p. 813-816
https://doi.org/10.5802/crmath.572
Research article / Article de recherche
Functional analysis / Analyse fonctionnelle
On the Scottish Book Problem 155 by Mazur
and Sternbach
Sur le problème 155 du Scottish Book par Mazur et
Sternbach
MichiyaMori a,b
a Graduate School of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba,
Meguro-ku, Tokyo, 153-8914, Japan
b Interdisciplinary Theoretical and Mathematical Sciences Program (iTHEMS),
RIKEN, 2-1 Hirosawa, Wako, Saitama, 351-0198, Japan
E-mail: mmori@ms.u-tokyo.ac.jp
Dedicated to Professor Yasuyuki Kawahigashi on the occasion of his 60+1st birthday.
Abstract. Problem 155 of the Scottish Book asks whether every bijection U : X → Y between two Banach
spaces X ,Y with the property that, each point of X has a neighborhood on which U is isometric, is globally
isometric on X . We prove that this is true under the additional assumption that X is separable and the weaker
assumption of surjectivity instead of bijectivity.
Résumé. Le problème 155 du Scottish Book demande si toute bijection U : X → Y entre deux espaces de
Banach X ,Y ayant la propriété que chaque point de X a un voisinage sur lequel U est isométrique, est
globalement isométrique sur X . Nous prouvons que ceci est vrai sous l’hypothèse supplémentaire que X
est séparable et l’hypothèse plus faible de surjectivité au lieu de bijectivité.
2020 Mathematics Subject Classification. 46B04.
Funding. The author is supported by JSPS KAKENHI Grant Number 22K13934.
Manuscript received 14 August 2023, revised 17 September 2023, accepted 18 September 2023.
According to [2], Problem 155 of the famous Scottish Book posed by Mazur and Sternbach on
November 18 in 1936 asks the following.
Given are two spaces X ,Y of type (B), y = U (x) is a one-to-one mapping of the
space X onto the whole space Y with the following property: For every x0 ∈ X
there exists an ε> 0 such that the mapping y =U (x), considered for x belonging
to the sphere with the center x0 and radius ε, is an isometric mapping. Is the
mapping y =U (x) an isometric transformation?
Let us rephrase this problem in a modern manner. For a point x0 in a Banach space X and a
positive real number r , we use the symbol B(x0,r ) := {x ∈ X | ∥x −x0∥ ≤ r } for the ball with center
x0 and radius r . Note that a sphere in the Scottish Book usually means a ball in this sense. (A set
of the form S(x0,r ) := {x ∈ X | ∥x − x0∥ = r } is called the surface of a sphere in the Scottish Book.
If we interpret that the “sphere” in the above problem means S(x0,r ), then one may encounter
simple counterexamples, see Remark 4.)
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
814 Michiya Mori
In what follows, we assume that X ,Y are two real Banach spaces (= spaces of type (B)). A
mapping f : D → E between subsets D ⊂ X , E ⊂ Y is called an isometry if it satisfies ∥ f (x1)−
f (x2)∥ = ∥x1 −x2∥ for any x1, x2 ∈ D . The above problem translates into the following.
Let U : X → Y be a bijection with the following property: For every x0 ∈ X there
exists an ε > 0 such that the mapping U restricted to B(x0,ε) is an isometry. Is
U : X → Y an isometry on X ?
One may find no commentaries on Problem 155 in the second printed edition of the Scottish
Book [2] published in 2015. It seems that the solution to this problem is still unknown. The pur-
pose of this note is to give a positive solution to this problem under the additional assumption
that X is separable and the weaker assumption of surjectivity instead of bijectivity. The nonsep-
arable case remains open.
The celebrated Mazur–Ulam theorem [3] asserts that every surjective isometry between two
real Banach spaces is affine. A version of the Mazur–Ulam theorem due to Mankiewicz [1]
asserts that any surjective isometry between balls of Banach spaces extends uniquely to an affine
surjective isometry of the whole Banach spaces. We give lemmas that are relevant to the Mazur–
Ulam and Mankiewicz’s theorems.
Lemma 1. Let x0 ∈ X and r > 0. Let a subset D ⊂ X satisfy B(x0,r ) ⊂ D. If g : D → X is an isometry
satisfying g (x) = x for every x ∈ B(x0,r ), then g (x) = x holds for every x ∈ D ∩B(x0,2r ).
Proof. We imitate the argument in the original proof of the Mazur–Ulam theorem [3]. For a
bounded subset C ⊂ X , we define δ(C ) := sup{∥x − x ′∥ | x, x ′ ∈ C }. Let x1 ∈ D ∩B(x0,2r ). Then
C1 := B(x0,∥x1 − x0∥/2)∩B(x1,∥x1 − x0∥/2) is a bounded subset with C1 ⊂ B(x0,r ). For n ≥ 1, we
inductively define Cn+1 := {x ∈ Cn | Cn ⊂ B(x,δ(Cn)/2)}. Then it clearly follows that C1 ⊃ C2 ⊃ ·· ·
and δ(Cn) ≤ 2−1δ(Cn−1) ≤ ·· · ≤ 2−n+1δ(C1). Therefore, we see that ⋂n≥1 Cn consists of at most
one point. We show that {x2} =⋂n≥1 Cn , where x2 = (x0 + x1)/2. By symmetricity, for every n, we
find that a point x lies in Cn if and only if 2x2 − x ∈Cn . It is clear that x2 ∈C1. Assume that n ≥ 1
and x2 ∈ Cn . If x ∈ Cn , then 2x2 − x ∈ Cn , thus 2∥x2 − x∥ = ∥(2x2 − x)− x∥ ≤ δ(Cn). It follows that
x2 ∈Cn+1. Thus we obtain {x2} =⋂n≥1 Cn .
Since g is an isometry and g (x0) = x0, we see that ∥g (x1)−x0∥ = ∥g (x1)−g (x0)∥ = ∥x1−x0∥. Set
Ĉ1 := B(x0,∥x1 − x0∥/2)∩B(g (x1),∥x1 − x0∥/2) and Ĉn+1 := {x ∈ Ĉn | Ĉn ⊂ B(x,δ(Ĉn)/2)} for n ≥ 1.
Then exactly the same argument as in the preceding paragraph shows that {(x0 + g (x1))/2} =⋂
n≥1 Ĉn . Observe that our assumption implies Ĉn = g (Cn) for every n ≥ 1. It follows that
(x0 + g (x1))/2 = g (x2). Since x2 ∈ C1 ⊂ B(x0,r ), we obtain g (x2) = x2 = (x0 + x1)/2. Thus
g (x1) = x1. □
Lemma 2. Let x0 ∈ X and r > 0. If f : B(x0,r ) → Y is an isometry and f (B(x0,r )) has nonempty
interior, then f extends uniquely to an affine surjective isometry from X onto Y .
Proof. We may assume x0 = 0 ∈ X and r = 1 without loss of generality. Take y1 in the interior of
f (B(0,1)). Set x1 := f −1(y1). Since f is an isometry, there is r1 > 0 such that B(x1,r1) ⊂ B(0,1)
and f (B(x1,r1)) = B(y1,r1). It follows from Mankiewicz theorem that there is an affine surjective
isometry g : X → Y such that f = g on B(x1,r1). Then the mapping F := g−1 ◦ f : B(0,1) → X is an
isometry satisfying F (x) = x for every x ∈ B(x1,r1).
We consider the set D of all z ∈ B(0,1) with the property that there is some ρ > 0 such that
B(z,ρ) ⊂ B(0,1) and F (x) = x for every x ∈ B(z,ρ). Observe that x1 ∈ D . Assume that z ∈ D . Let
m be the smallest positive integer with B(z,2mρ) ̸⊂ B(0,1). Applying Lemma 1 (with center z)
repeatedly, we obtain F (x) = x for every x ∈ B(z,2mρ)∩B(0,1). In particular, for z =λz0 ∈ D with
∥z0∥ = 1 and 0 ≤ λ < 1, we see µz0 ∈ D for every µ ∈ (2λ−1,1). Beginning with z = x1 and using
this repeatedly, we obtain 0 ∈ D . Then another repeated application of Lemma 1 (with center 0)
shows that F (x) = x for every x ∈ B(0,1). □
Michiya Mori 815
Now we are ready to prove the main theorem of this note.
Theorem 3. Assume that X is separable. Let U : X → Y be a surjection with the following property:
For every x0 ∈ X there exists an εx0 > 0 such that the mapping U restricted to B(x0,εx0 ) is an
isometry. Then U : X → Y is an isometry on X .
Proof. Recall that a separable metric space is Lindelöf [4, Theorem 16.11] (every open cover has
countable subcover). Since the family of open sets {{x ∈ X | ∥x − x0∥ < εx0 } | x0 ∈ X } covers the
separable metric space X , one may find a sequence x1, x2, . . . in X such that X ⊂⋃n≥1{x ∈ X | ∥x−
xn∥ < εxn }. For each n, U restricts to an isometry on B(xn ,εxn ), and thus its image U (B(xn ,εxn )),
being complete, is closed in Y . Since U is surjective, we see that Y =⋃n≥1 U (B(xn ,εxn )). Thus Y
is written as a countable union of closed subsets. By the Baire Category Theorem, there is some
m ≥ 1 such that U (B(xm ,εxm )) has nonempty interior. Thus Lemma 2 implies that the restriction
of U to B(xm ,εxm ) extends to a surjective isometry f : X → Y .
Let us consider the nonempty open set D that consists of all z ∈ X with the property that
U (x) = f (x) for every x in some neighborhood of z. Assume that D ̸= X . Since X is connected,
we may take a boundary point x0 of D . Then U restricted to B(x0,εx0 ) is an isometry, and
B(x0,εx0 )∩D has nonempty interior. Since U = f on D and f is an affine surjective isometry,
we see from Lemma 2 that U = f on B(x0,εx0 ). This leads to x0 ∈ D , a contradiction. □
Remark 4. Let U : X → Y be a bijection with the following property: For every x0 ∈ X there exists
an ε> 0 such that the mapping U restricted to S(x0,ε) is an isometry. Is the mapping U : X → Y
an isometry on X ? The answer is NO. To see this, consider an arbitrary X and assume X = Y .
Let U : X → X be ANY mapping with the properties that U restricts to a bijection on B(0,1) and
that U (x) = x for every x ∈ X \ B(0,1). Even if U behaves very wildly on B(0,1), U clearly satisfies
the assumption of this question. Note that in the Scottish Book it is remarked that the answer
to Problem 155 is affirmative if we additionally assume U−1 is continuous, and this is automatic
when dimY < ∞ or Y is strictly convex. Therefore, the “sphere” in Problem 155 by no means
stands for the set S(x0,ε).
Remark 5. Can we drop the assumption of surjectivity in Theorem 3? The answer is NO. Indeed,
considering a piecewise linear mapping from R into the plane equipped with the ℓ1-norm such
that the velocity on each point is either ±(1,0) or ±(0,1), one may find many wild examples.
Acknowledgements
The author appreciates Kan Kitamura (University of Tokyo) for leading the author to the correct
interpretation of the term “sphere” in Problem 155 at the conference Operator Algebras and
Mathematical Physics in July 2023 at the University of Tokyo, which celebrates Professor Yasuyuki
Kawahigashi’s 60th birthday.
Declaration of interests
The authors do not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and have declared no affiliations other than their research
organizations.
References
[1] P. Mankiewicz, “On extension of isometries in normed linear spaces”, Bull. Acad. Polon. Sci.,
Sér. Sci. Math. Astron. Phys. 20 (1972), pp. 367–371.
816 Michiya Mori
[2] R. D. Mauldin, The Scottish Book. Mathematics from the Scottish Café. With selected problems
from the New Scottish Book, 2nd updated and enlarged edition, Birkhäuser/Springer, 2015.
[3] S. Mazur and S. Ulam, “Sur les transformationes isométriques d’espaces vectoriels normés”,
C. R. Acad. Sci. Paris 194 (1932), pp. 946–948.
[4] S. Willard, General Topology, Dover Publications, Mineola, 2004.


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